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v1 [

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Jun

2012

The 2-page crossing number of Kn

Bernardo M. AbregoCalifornia State University, Northridge

bernardo.abrego@csun.edu

Oswin AichholzerGraz University of Technology

oaich@ist.tugraz.at

Silvia Fernandez-MerchantCalifornia State University, Northridge

silvia.fernandez@csun.edu

Pedro RamosUniversidad de Alcala

pedro.ramos@uah.es

Gelasio SalazarUniversidad Autonoma de San Luis Potosı

gsalazar@ifisica.uaslp.mx

August 6, 2018

Abstract

Around 1958, Hill described how to draw the complete graph Kn with

Z (n) :=1

4

⌊

n

2

⌋

⌊

n− 1

2

⌋⌊

n− 2

2

⌋ ⌊

n− 3

2

⌋

crossings, and conjectured that the crossing number cr(Kn) of Kn is exactly Z(n).This is also known as Guy’s conjecture as he later popularized it. Towards the endof the century, substantially different drawings of Kn with Z(n) crossings were found.These drawings are 2-page book drawings, that is, drawings where all the vertices areon a line ℓ (the spine) and each edge is fully contained in one of the two half-planes(pages) defined by ℓ. The 2-page crossing number of Kn, denoted by ν2(Kn), is theminimum number of crossings determined by a 2-page book drawing of Kn. Sincecr(Kn) ≤ ν2(Kn) and ν2(Kn) ≤ Z(n), a natural step towards Hill’s Conjecture is theweaker conjecture ν2(Kn) = Z(n), popularized by Vrt’o. In this paper we develop anovel and innovative technique to investigate crossings in drawings of Kn, and use itto prove that ν2(Kn) = Z(n). To this end, we extend the inherent geometric definitionof k-edges for finite sets of points in the plane to topological drawings of Kn. We alsointroduce the concept of ≤≤k-edges as a useful generalization of ≤k-edges and extenda powerful theorem that expresses the number of crossings in a rectilinear drawing of

1

Kn in terms of its number of (≤ k)-edges to the topological setting. Finally, we give acomplete characterization of crossing minimal 2-page book drawings of Kn and showthat, up to equivalence, they are unique for n even, but that there exist an exponentialnumber of non-homeomorphic such drawings for n odd.

1 Introduction

In a drawing of a graph in the plane, each vertex is represented by a point and each edge isrepresented by a simple open arc, such that if uv is an edge, then the closure (in the plane)of the arc α representing uv consists precisely of α and the points representing u and v. Itis further required that no arc representing an edge contains a point representing a vertex.

A crossing in a drawing D of a graph G is a pair (x, {α, β}), where x is a point in theplane, α, β are arcs representing different edges, and x ∈ α ∩ β. The crossing number cr(D)of D is the number of crossings in D, and the crossing number cr(G) of G is the minimumcr(D), taken over all drawings D of G.

A drawing is good if (i) no three distinct arcs representing edges meet at a common point;(ii) if two edges are adjacent, then the arcs representing them do not intersect each other;and (iii) an intersection point between two arcs representing edges is a crossing rather thantangential. It is well-known (and easy to prove) that every graph has a crossing-minimaldrawing which is good (moreover, (ii) and (iii) hold in every crossing-minimal drawing).Thus, when our aim (as in this paper) is to estimate the crossing number of a graph, we mayassume that all drawings under consideration are good.

As usual, for simplicity we often make no distinction between a vertex and the pointrepresenting it, or between an edge and the arc representing it. No confusion should arisefrom this practice.

Around 1958, Hill conjectured that

cr(Kn) = Z(n) :=1

4

⌊n

2

⌋

⌊

n− 1

2

⌋⌊

n− 2

2

⌋⌊

n− 3

2

⌋

. (1)

This conjecture appeared in print a few years later in papers by Guy [15] and Harary andHill [16]. Hill described drawings of Kn with Z(n) crossings, which were later corroboratedby Blazek and Koman [5]. These drawings show that cr(Kn) ≤ Z(n). The best knowngeneral lower bound is limn→∞ cr(Kn)/Z(n) ≥ 0.8594, due to de Klerk et al. [11]. For moreon the history of this problem we refer the reader to the excellent survey by Beineke andWilson [3].

One of the major motivations for investigating crossing numbers is their application toVLSI design. With this motivation in mind, Chung, Leighton and Rosenberg [7] analyzed

2

embeddings of graphs in books: the vertices lie on a line (the spine) and the edges lie on thepages of the book. Book embeddings of graphs have been extensively studied [4, 12]. Nowif the book has k pages, and crossings among edges are allowed, the result is a k-page bookdrawing.

Here we concentrate on 2-page book drawings. The 2-page crossing number ν2(G) of agraph G is the minimum of cr(D) taken over all 2-page book drawings D of G. Alternativeterminologies for the 2-page crossing number are circular crossing number [17] and fixedlinear crossing number [8]. We may regard the pages as the closed half-planes defined bythe spine, and so every 2-page book drawing can be realized as a plane drawing; it followsthat cr(G) ≤ ν2(G) for every graph G.

In 1964, Blazek and Koman [5] found 2-page book drawings of Kn with Z(n) crossings,thus showing that ν2(Kn) ≤ Z(n) (see also Guy et al. [14], Damiani et al. [9], Harborth[17], and Shahrokhi et al. [20].) Once these constructions were known, the conjecture thatν2(Kn) = Z(n) is implicit in the conjecture given by Equation (1) since cr(Kn) ≤ ν2(Kn).However, the only explicit reference to this weaker conjecture is, as far as we know, fromVrt’o [21].

Buchheim and Zhang [6] reformulated the problem of finding ν2(Kn) as a maximum cutproblem on associated graphs, and then solved exactly this maximum cut problem for alln ≤ 13, thus confirming Equation (1) for 2-page book drawings for all n ≤ 14 (the casen = 14 follows from the case n = 13 by an elementary counting argument). Very recently,De Klerk and Pasechnik [10] used this max cut reformulation to find the exact value of ν2(Kn)for all n ≤ 21 and n = 24, and moreover, by using semidefinite programming techniques, toobtain the asymptotic bound limn→∞ ν2(Kn)/Z(n) ≥ 0.9253. All the results reported in [6]and [10] are computer-aided.

In this paper we prove that ν2(Kn) = Z(n). The main technique for the proof is the exten-sion of the concept of k-edge of a finite set of points to topological drawings of the completegraph. We do this in a way such that the identities proved by Abrego and Fernandez-Merchant [1] and Lovasz et al. [19], that express the crossing number of a rectilinear drawingof Kn in terms of the k-edges or the (≤ k)-edges of its set of vertices, are also valid in thetopological setting.

We recall that a drawing D is rectilinear if the edges of D are straight line segments, andthe rectilinear crossing number cr(G) of a graph G is the minimum of cr(D) taken over allrectilinear drawings D of G. An edge pq of D is a k-edge if the line spanned by pq dividesthe remaining set of vertices into two subsets of cardinality k and n− 2− k. Thus a k-edgeis also an (n − 2 − k)-edge. Denote by Ek(D) the number of k-edges of D. The followingidentity [1, 19] has been key to the recent developments on the rectilinear crossing numberof Kn.

3

cr (D) = 3

(

n

4

)

−

⌊n/2⌋−1∑

k=0

k (n− 2− k)Ek (D) . (2)

In Section 2 we generalize the concept of k-edge to arbitrary (that is, not necessarilyrectilinear) drawings of Kn. This allows us to extend Equation (2) to (good) topologicaldrawings of Kn. The key observation to extend the definition of k-edge to the new setting isto observe that, although half-planes are not well defined, we can use the orientation of thetriangles defined by three points: the edge pq will be a k-edge of the topological drawing ifthe set of triangles adjacent to pq is divided, according to its orientation, into two subsetswith cardinality k and n−k−2. In Section 3 we use this tool to show that ν2(Kn) = Z(n). Inorder to do that, we need to introduce the new concept of ≤≤k-edges, because for topologicaldrawings the lower bound for ≤k-edges, E≤k(D) ≥ 3

(

k+22

)

does not hold. In Section 4 weanalyze crossing optimal 2-page drawings of Kn. We give a complete characterization oftheir structure, showing that, up to equivalence (see Section 4.1 for a detailed definition),crossing optimal drawings are unique for n even. In contrast, for n odd we provide a familyof size 2(n−5)/2 of non-equivalent crossing optimal drawings. We conclude with some openquestions and directions for future research in Section 5.

An extended abstract of this paper [2] has appeared. In it we include some additionalobservations on the structure of crossing optimal 2-page drawings of Kn. For instance, inthese drawings the above mentioned inequality E≤k (D) ≥ 3

(

k+22

)

does hold.

2 Crossings and k-edges

In this section we generalize the concept of k-edges, which has so far only been used in thegeometric setting of finite sets of points in the plane, to topological drawings of Kn. LetD be a good drawing of Kn, let

−→pq be a directed edge of D, and r a vertex of D otherthan p or q. We say that r is on the left (respectively, right) side of −→pq if the topologicaltriangle pqr traced in that order (its vertices and edges correspond to those in D) is orientedcounterclockwise (respectively, clockwise). Note that this is well defined as the three edgespq, qr, and rp in D do not self intersect and do not intersect each other, since D is good.We say that the edge pq is a k-edge of D if it has exactly k points of D on the same side (leftor right), and thus n− 2− k points on the other side. Hence, as in the geometric setting, ak-edge is also an (n−2−k)-edge. Note that the direction of the edge pq is no longer relevantand every edge of D is a k-edge for some unique k such that 0 ≤ k ≤ ⌊n/2⌋ − 1. Let Ek(D)be the number of k-edges of D.

Theorem 1. For any good drawing D of Kn in the plane the following identity holds,

cr (D) = 3

(

n

4

)

−

⌊n/2⌋−1∑

k=0

k (n− 2− k)Ek (D) .

4

A B C

Figure 1: The three different good drawings of K4, with 3, 2, and 2 separations. The edgeof each separation is shown bold.

Proof. In a good drawing of Kn, we say that an edge pq separates the vertices r and s ifthe orientations of the triangles pqr and pqs are opposite. In this case, we say that theset {pq, r, s} is a separation. It is straightforward to check that, up to ambient isotopyequivalence, there are only three different good drawings A,B,C of K4; these are shown inFigure 1.

We denote by TA, TB, and TC the number of induced subdrawings of D of type A, B, andC, respectively. Then

TA + TB + TC =

(

n

4

)

, (3)

and since the subdrawings of types B or C are in one-to-one correspondence with the cross-ings of D, it follows that

cr (D) = TB + TC . (4)

We count the number of separations in D in two different ways: First, each subdrawingof type A has 3 separations (the edge in each separation is bold in Figure 1), and eachsubdrawing of types B or C has 2 separations. This gives a total of 3TA + 2TB + 2TC

separations in D. Second, each k-edge belongs to exactly k(n−2−k) separations. Summing

over all k-edges for 0 ≤ k ≤ ⌊n/2⌋−1 gives a total of∑⌊n/2⌋−1

k=0 k(n−2−k)Ek(D) separationsin D. Therefore

3TA + 2TB + 2TC =

⌊n/2⌋−1∑

k=0

k (n− 2− k)Ek(D). (5)

Finally, subtracting Equation (5) from three times Equation (3) we get

TB + TC = 3

(

n

4

)

−

⌊n/2⌋−1∑

k=0

k (n− 2− k)Ek(D),

and thus by Equation (4) we obtain the claimed result.

For 0 ≤ k ≤ ⌊n/2⌋ − 1 and D a good drawing of Kn, we define the set of ≤k-edges of Das all j-edges in D for j = 0, . . . , k. The number of ≤k-edges of D is denoted by

E≤k (D) :=

k∑

j=0

Ej (D) .

5

Similarly, we denote the number of ≤≤k-edges of D by

E≤≤k (D) :=k

∑

j=0

E≤j (D) =k

∑

j=0

j∑

i=0

Ei (D) =k

∑

i=0

(k + 1− i)Ei (D) .

To avoid special cases we define E≤≤−1(D) = E≤≤−2(D) = 0.

The following result restates Theorem 1 in terms of the number of ≤≤k-edges.

Proposition 2. Let D be a good drawing of Kn. Then

cr(D) = 2

⌊n/2⌋−2∑

k=0

E≤≤k(D)−1

2

(

n

2

)⌊

n− 2

2

⌋

−1

2(1 + (−1)n)E≤≤⌊n/2⌋−2(D).

Proof. First note that for 2 ≤ k ≤ ⌊n/2⌋−1 we have that E≤≤k(D)−E≤≤k−1(D) = E≤k(D)and E≤k(D)− E≤k−1(D) = Ek(D). Thus

Ek (D) = E≤≤k (D)− 2E≤≤k−1 (D) + E≤≤k−2 (D) .

We rewrite the last term in Theorem 1.

⌊n/2⌋−1∑

k=0

k(n− 2− k)Ek(D)

=

⌊n/2⌋−1∑

k=2

k(n− 2− k)[E≤≤k(D)− 2E≤≤k−1(D) + E≤≤k−2(D)]

=

⌊n/2⌋−3∑

k=0

(k(n− 2− k)− 2(k + 1)(n− 3− k) + (k + 2)(n− 4− k))E≤≤k(D)

+(⌊n

2

⌋

− 1)(

n− 1−⌊n

2

⌋)

E≤≤⌊n/2⌋−1(D) + (−2(⌊n

2

⌋

− 1)(

n− 1−⌊n

2

⌋)

+(⌊n

2

⌋

− 2)(

n−⌊n

2

⌋)

)E≤≤⌊n/2⌋−2(D)

= −2

⌊n/2⌋−3∑

k=0

E≤≤k(D) +(⌊n

2

⌋

− 1)(

n− 1−⌊n

2

⌋)

E≤≤⌊n/2⌋−1(D)

+ (−2(⌊n

2

⌋

− 1)(

n− 1−⌊n

2

⌋)

+(⌊n

2

⌋

− 2)(

n−⌊n

2

⌋)

)E≤≤⌊n/2⌋−2(D).

Since E≤≤⌊n/2⌋−1(D) = E≤≤⌊n/2⌋−2(D) + E≤⌊n/2⌋−1(D) = E≤≤⌊n/2⌋−2(D) +(

n2

)

, it follows by

6

Theorem 1 that

cr (D) = 3

(

n

4

)

−

⌊n/2⌋−1∑

k=0

k (n− 2− k)Ek (D) = 3

(

n

4

)

+ 2

⌊n/2⌋−3∑

k=0

E≤≤k (D)

+(

n+ 1− 2⌊n

2

⌋)

E≤≤⌊n/2⌋−2(D)−(⌊n

2

⌋

− 1)(

n− 1−⌊n

2

⌋)

(

n

2

)

= 2

⌊n/2⌋−3∑

k=0

E≤≤k (D)−1

2

(

n

2

)⌊

n− 2

2

⌋

+

{

E≤≤⌊n/2⌋−2(D) if n is even,

2E≤≤⌊n/2⌋−2(D) if n is odd,

which is equivalent to the claimed result.

3 The 2-page crossing number

We are concerned with 2-page book drawings of Kn. Obviously any line can be chosen as thespine, and for the rest of the paper we will assume that the spine is the x-axis. Moreover,by topological equivalence, we will assume that the vertices are precisely the points withcoordinates (1, 0), (2, 0), . . . , (n, 0).

Consider a 2-page book drawing D of Kn, and label the vertices 1, 2, . . . , n from left toright. Our interest lies in crossing optimal drawings, and it is readily seen that in every suchdrawing, none of the edges (1, 2), (2, 3), . . . , (n− 1, n), (n, 1) is crossed. Thus we may chooseto place each of these edges in either the upper closed halfplane (page) or in the lower closedhalfplane (page). Moreover, we may choose to place each of the edges (1, 2), (2, 3), . . . , (n−1, n) completely on the spine, and this is the convention we shall adopt for the rest of thepaper. The edge (n, 1) may be placed indistinctly in the upper page or in the lower page,and for the rest of the paper we adopt the convention that it is place in the upper page.Moreover, because we are only concerned with good drawings, we assume without loss ofgenerality that the rest of the edges are semicircles.

Color the edges above or on the spine blue and below the spine red, respectively. Weconstruct an upper triangular matrix which corresponds to the coloring of these edges, seeFigure 2. We call this the 2-page matrix of D and denote it by M (D). Label the columnsof the 2-page matrix with 2, . . . , n from left to right and the rows with 1, 2, . . . , n − 1 fromtop to bottom. For i < j an entry (i, j) (row,column) in the 2-page matrix M(D) is a pointwith the same color as the edge ij in the drawing D.

Remark. It follows from the convention laid out above that for every 2-page book drawingD, the entries (1, 2), (2, 3), . . . , (n− 1, n) and (1, n) in M(D) are all blue.

We start by proving some basic properties of the 2-page matrix.

7

1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

Figure 2: Two-colored diagram for a 2-page book drawing D of K8 and the corresponding2-page matrix M(D). Solid dots and lines represent blue edges. Open dots and dashed linesrepresent red edges. From our convention to place the edges (1, 2), (2, 3), . . . , (n − 1, n) onthe spine and the edge (1, n) in the upper page, it follows that all the entries in the maindiagonal, as well as the upper right corner entry, are blue.

Lemma 3. Let D be a 2-page book drawing of Kn. For 1 ≤ i < j ≤ n, let k be the sumof the number of points to the right plus the number of points above the entry (i, j) in the2-page matrix of D, which have the same color as (i, j). Then the edge ij is a k-edge. (It ispossible to have k > ⌊n/2⌋ − 1.)

Proof. Let 1 ≤ i < j ≤ n and assume that the edge ij is blue (red). We count the numberof points l in D to the left (right) of ij. For l 6∈ {i, j} the triangle ijl is oriented counter-clockwise (clockwise) if and only if either l < i and the edge lj is blue (red), or l > j andthe edge il is blue (red). In the first case these edges correspond to blue (red) points abovethe entry (i, j), and in the second case to blue (red) points to the right of the entry (i, j),respectively.

In view of Lemma 3 we say that the point in the entry (i, j) of the 2-page matrix of Drepresents a k-edge if ij is a k-edge (or an (n− 2− k)-edge) in D.

Lemma 4. For k < n/2 − 1 and for 1 ≤ j ≤ k + 1, in the 2-page matrix of a drawing Dof Kn there are at least 2 (k + 2− j) points in row j representing ≤k-edges. Similarly, forn− k ≤ j ≤ n there are at least 2 (k + 1− n + j) points in column j representing ≤k-edges.

Proof. For 1 ≤ j ≤ k + 1, in row j the rightmost k + 2 − j points of each color represent≤k-edges as they have at most k + 1− j points of their color to the right and at most j − 1on top. So if each color appears at least k + 2 − j times in row j, we have guaranteed2 (k + 2− j) ≤k-edges in row j. If one of the colors appears fewer than k + 2 − j times, sothat there are k + 2− j − e blue points in row j for some 1 ≤ e ≤ k + 2− j, then there are

8

n − j − (k + 2− j − e) = n − 2 − k + e red points in this row. In this case we claim thatalso the leftmost e red points in this row represent ≤k-edges. In fact, for 1 ≤ i ≤ e, the i-thred point (from the left) in row j, has exactly n− 2 − k + e− i red points to the right andperhaps more red points on top. Since for n ≥ 2 we have n− 2 − k + e− i ≥ n/2 − k, thisi-th red point also represents a ≤k-edge. The equivalent result holds for the rightmost k+1columns.

Lemma 5. For 0 ≤ j < n/2− 1, in the 2-page matrix of a drawing D of Kn there are twopoints in column n which correspond to j-edges in D. For n even there exists one such pointin column n corresponding to an (n/2− 1)-edge in D.

Proof. We follow the lines of the proof of Lemma 4. Consider the points in column n in orderfrom top to bottom. By Lemma 3 the i-th vertex of a color corresponds to an (i− 1)-edge.Thus, if there are at least j+1 vertices for each color we are done. Otherwise assume withoutloss of generality that there are j + 1 − e blue points in column n for some 1 ≤ e ≤ j + 1.Then there are n−1−(j+1−e) = n−j+e−2 red points in this column. For 1 ≤ i ≤ ⌊n/2⌋the i-th red point corresponds to an (i − 1)-edge, and for ⌊n/2⌋ + 1 ≤ i ≤ n − j + e − 2the i-th red point corresponds to an (i− 1) = (n− i− 1)-edge. Thus we get two red pointscorresponding to j-edges for i = j+1 and i = n−j−1. Finally, observe that these two pointsare different for j < n/2− 1. For n even we get only one such point for j = n/2− 1.

The next theorem gives a lower bound on the number of ≤≤k-edges, which will play acentral role in deriving our main result. We need the following definitions. Let D be a gooddrawing of Kn. Let l be a vertex of Kn, and let D′ be the (evidently, also good) drawing ofKn−1 obtained by deleting from D the vertex l and its adjacent edges. Note that a k-edge ijin D′ is a k-edge or a (k + 1)-edge in D. Indeed, if ij has exactly k points to its right in D′

(an equivalent argument holds if the k points are on its left), then there are k or k+1 pointsto the right of ij in D depending on whether l is to the left or to the right, respectively, ofij. We say that a k-edge in D is (D,D′)-invariant if it is also a k-edge in D′. Wheneverit is clear what D and D′ are, we simply say that an edge is invariant. A (D,D′)-invariant≤ k-edge is a (D,D′)-invariant j-edge for some 0 ≤ j ≤ k ≤ n/2− 1. Denote by E≤k(D,D′)the number of (D,D′)-invariant ≤ k-edges.

Theorem 6. Let n ≥ 3. For every 2-page book drawing D of Kn and 0 ≤ k < n/2 − 1, wehave

E≤≤k (D) ≥ 3(

k+33

)

.

Proof. We proceed by induction on n. The induction base n = 3 holds trivially. For n ≥ 4,consider a 2-page book drawing D of Kn with horizontal spine and label the vertices fromleft to right with 1, 2, . . . , n. Remove the point n and all incident edges to obtain a 2-pagebook drawing D′ of Kn−1. To bound E≤≤k (D), recall that

E≤≤k (D) =

k∑

j=0

(k + 1− j)Ej (D) . (6)

9

All edges incident to n are in D but are not in D′. In fact, by Lemma 5, there are twoj-edges adjacent to the vertex n for each 0 ≤ j ≤ k ≤ ⌊n/2⌋ − 2. These edges contributewith 2

∑kj=0(k + 1− j) = 2

(

k+22

)

to Equation (6). We next compare Equation (6) to

E≤≤k−1 (D′) =

k−1∑

j=0

(k − j)Ej (D′) . (7)

Any edge contributing to Equation (7) also contributes to Equation (6), but possibly witha different value. As observed before, a j-edge in D′ is a j-edge or a (j + 1)-edge in D.A j-edge in D′ contributes to Equation (7) with k − j. A j-edge and a (j + 1)-edge in Dcontribute to Equation (6) with k + 1 − j and k − j, respectively. This is a gain of +1 or0, respectively, towards E≤≤k(D) when compared to E≤≤k−1(D

′). Finally, a k-edge in bothD and D′ does not contribute to Equation (7) and contributes to Equation (6) with +1.Therefore

E≤≤k(D) = E≤≤k−1(D′) + 2

(

k + 2

2

)

+ E≤k(D,D′).

By induction hypothesis, E≤≤k−1(D′) ≥ 3

(

k+23

)

and thus

E≤≤k(D) ≥ 3

(

k + 2

3

)

+ 2

(

k + 2

2

)

+ E≤k(D,D′) = 3

(

k + 3

3

)

−

(

k + 2

2

)

+ E≤k(D,D′).

We finally prove that

E≤k(D,D′) ≥

(

k + 2

2

)

. (8)

In fact, we prove that for each 1 ≤ j ≤ k + 1 there are at least k + 2 − j points in row jof M(D) that represent (D,D′)-invariant ≤ k-edges. Suppose that the edge jn is blue (theequivalent argument holds when jn is red). Then any red point in row j with i ≤ k redpoints above or to its right inM(D) represents a (D,D′)-invariant i-edge; and any blue pointin row j with i ≥ n − 2 − k blue points above or to its right represents a (D,D′)-invariant(n− 2 − i)-edge. Thus, the first k + 2− j red points from the right in row j (if they exist)represent (D,D′)-invariant ≤ k-edges as they have at most k + 2 − j − 1 red points to theright and at most j − 1 red points above in both M (D) and M (D′). If there are fewerthan k + 2 − j red points in row j of M(D), say k + 2 − j − e for some 1 ≤ e ≤ k + 2 − j,then the first e blue points in row j of M(D) from the left represent ≤k-edges, because theyhave at least n − j − e ≥ n − j − k − 2 + j = n − k − 2 blue points to their right. Hencethere are at least k + 2 − j − e red points and at least e blue points (for a total of at leastk + 2 − j points) that represent (D,D′)-invariant ≤ k-edges in row j of M(D). Summingover all 1 ≤ j ≤ k + 1, we get that

E≤k(D,D′) ≥

k+1∑

j=1

(k + 2− j) =

(

k + 2

2

)

.

We are now ready to prove our main result, namely that the 2-page crossing number ofKn is Z(n).

10

Theorem 7. For every positive integer n, ν2(Kn) = Z(n).

Proof. The cases n = 1 and n = 2 are trivial. Let n ≥ 3. As we mentioned above, 2-pagebook drawings with Z (n) crossings were constructed by Blazek and Koman [5] (see also Guyet al. [14], Damiani et al. [9], Harborth [17], and Shahrokhi et al. [20].) These drawings showthat ν2 (Kn) ≤ Z (n). For the lower bound, let D be a 2-page book drawing of Kn. UsingProposition 2 and Theorem 6, we obtain

cr (D) ≥ 2

⌊n/2⌋−2∑

k=0

3

(

k + 3

3

)

−1

2

(

n

2

)⌊

n− 2

2

⌋

−3

2(1 + (−1)n)

(⌊

n2

⌋

+ 1

3

)

= 6

(⌊

n2

⌋

+ 2

4

)

−1

2

(

n

2

)⌊

n− 2

2

⌋

−3

2(1 + (−1)n)

(⌊

n2

⌋

+ 1

3

)

=

{

164(n− 1)2 (n− 3)2 if n is odd,

164n (n− 2)2 (n− 4) if n is even,

= Z(n).

4 Crossing optimal configurations

In all this section D denotes a 2-page book drawing of Kn and M(D) its 2-page matrix. Wesay that D is crossing optimal if ν2(D) = Z(n). Theorem 19 in Subsection 4.3 describesthe general structure of the crossing optimal 2-page book drawings of Kn. We use it toprove that, up to the equivalence described below, there is a unique crossing optimal 2-pagebook drawing of Kn when n is even and, in contrast, there exists an exponential number ofnon-equivalent crossing optimal 2-page book drawings of Kn when n is odd.

4.1 Equivalent drawings

Let D be a 2-page book drawing of Kn. Recall that we are assuming that the vertices ofD are the points {(i, 0) : 1 ≤ i ≤ n}. Consider the following transformation f that resultsin the 2-page book drawing f (D) of Kn: move the vertex (1, 0) to the point (n, 0), and forevery 2 ≤ k ≤ n move the vertex (k, 0) to the vertex (k − 1, 0). That is, if an edge 1j wasdrawn above (below) the spine in D, then the edge (j − 1)(n) is drawn above (below) thespine in f (D); for all other edges ij with 1 < i < j ≤ n, if ij was drawn above (below)the spine in D, then the edge (i − 1)(j − 1) is drawn above (below) the spine in f (D).Note that D and f (D) have the same number of crossings, and fn(D) = D. There aretwo other natural transformations of a drawing D: A vertical reflection g(D) about the linewith equation x = n/2 and a horizontal reflection h(D) about the spine (or x-axis). In g(D)an edge ij is drawn above (below) the spine if the edge (n + 1 − j)(n + 1 − i) is drawnabove (below) the spine in D. In h(D) an edge ij is drawn above (below) the spine if the

11

Figure 3: A 2-page drawing of K8 and its strip diagram.

edge ij is drawn below (above) the spine in D. Note that g2(D) = h2(D) = D. Given a2-page drawing D, all drawings obtained by compositions of these transformations from Dare said to be equivalent to D. All drawings obtained this way are topologically isomorphic(homeomorphic) and thus they all have the same number of crossings as D. The groupspanned by these transformations is isomorphic to the direct sum of the dihedral groupD2n and the group with 2 elements Z2. The set {f, g, h} is a set of generators such thatg2 = h2 = fn = 1, g◦f = f−1◦g, h◦f = f ◦h, and g◦h = h◦g. Thus the 4n transformationsin the group can be parametrized by ha ◦ gb ◦ f i with i ∈ {0, 1, . . . , n− 1} and a, b ∈ {0, 1}.

Now we describe these transformations in the 2-page matrix diagram of D: To obtainM (f (D)) from M (D), we simply rotate 90 degrees counterclockwise the first row of M (D)and use it as the nth column of M (f (D)). The diagram M (g (D)) is obtained from M (D)by reflecting with respect to the diagonal {(i, n+ 1− i) : 1 ≤ i ≤ ⌊n/2⌋}. Finally, M(h(D))is obtained by switching the color of every point except those that join consecutive verticeson the spine or the point (1, n). We can place M (D) and M (f (D)) together so that thepart they have in common overlaps. Doing this for M (fm (D)) for all integers m we obtaina periodic double infinite strip with period n and with a horizontal section that is n − 1units wide. We call this the strip diagram of D, or of fm (D) for any integer m. (SeeFigure 3.) Any right triangular region with the same dimensions as M (D) obtained fromthe strip diagram of D by a horizontal and a vertical cut is the matrix diagram of a drawingequivalent to D and thus it has the same number of crossings as D.

4.2 Properties of crossing optimal drawings

We start with a couple of definitions. Consider the entry (i, j) of M(D). We order the entriesin row i to the left of (i, j) as follows: first all entries, from right to left, whose color differsto that of (i, j), followed by all other entries (those with the same color as (i, j)) from left to

12

right. This is called the order associated to (i, j). Observe that this is the order in which theedges il (i < l < j) appear in the 2-page drawing, ordered bottom to top if the edge ij is blueand top to bottom if the edge ij is red. Let c be an integer such that 0 ≤ c ≤ n− 1. Denoteby Dc the subgraph of D obtained by deleting the c right-most points of D, or equivalently,M(Dc) is obtained by deleting the last c columns of M (D). The following results stronglyrely on the proof of Theorem 6.

Lemma 8. Suppose that l ≥ i+m+1 for some integers 1 ≤ i < l < j ≤ n and 1 ≤ m < j−i.The entry (i, l) is one of the first m entries in the order associated to (i, j) if and only if(i, l) and (i, j) have different colors.

Proof. Note that if (i, l) and (i, j) have the same color, then all entries to the left of (i, l)come before (i, l) in the order associated to (i, j).

Lemma 9. Let p be an integer such that 0 ≤ p ≤ ⌊n/2⌋−2. Suppose that E≤k(D,D1) =(

k+22

)

for all 0 ≤ k ≤ p. Then M (D) satisfies that for 1 ≤ i ≤ p + 1 in row i there is exactlyone (D,D1)-invariant k-edge for each i − 1 ≤ k ≤ p, and there are no (D,D1)-invariant(≤ i− 2)-edges. In all other rows there are no (D,D1)-invariant ≤ p-edges.

Proof. In what follows all invariant edges are (D,D1)-invariant edges. For k = 0 the state-ment implies that there is a unique invariant 0-edge and it appears in row 1. Note that thisedge corresponds to the first entry in the order associated to (1, n) in M(D). Following theproof of Theorem 6, E≤k(D,D1) =

(

k+22

)

implies that for all 1 ≤ i ≤ k + 1 there are exactlyk + 2 − i invariant ≤ k-edges in row i of M (D), and for k + 2 ≤ i ≤ n − 1 there are noinvariant ≤ k-edges in row i of M(D). The second part implies that there are no invariantk-edges in row i for all k + 2 ≤ i ≤ n − 1 and 0 ≤ k ≤ p. Similarly, E≤k−1(D,D1) =

(

k+12

)

implies that for all 1 ≤ i ≤ k there are exactly k + 1 − i invariant (≤ k − 1)-edges in row iof M (D). Therefore for all 1 ≤ i ≤ k there is exactly (k + 2− i)− (k + 1− i) = 1 invariantk-edge, and for i = k + 1 there is exactly k + 2 − (k + 1) = 1 invariant ≤ k-edge and noinvariant (≤ k − 1)-edge in row i of M (D). Therefore, there is exactly one invariant k-edgein row k + 1.

Lemma 10. Let p be an integer such that 0 ≤ p ≤ ⌊n/2⌋ − 2.

i) Suppose that for some 1 ≤ i ≤ p+ 1 row i of M(D) has exactly one (D,D1)-invariantk-edge for each i− 1 ≤ k ≤ p and no (D,D1)-invariant ≤ (i− 2)-edges. If the entry (i, n) inM(D) is blue (red), then the mth entry in row i in the order associated to (i, n) has at leastmin{p+ 2−m, i− 1} red (blue) entries above for every 1 ≤ m ≤ min{p+ 1, n− i− 1}.

ii) Suppose that for some i ≥ p + 2 row i of M(D) does not have (D,D1)-invariant≤ p-edges. If the entry (i, n) in M(D) is blue (red), then the mth entry in row i in theorder associated to (i, n) has at least p + 2 −m red (blue) entries above for every 1 ≤ m ≤min{p+ 1, n− i− 1}.

13

Proof. In what follows invariant edges refer to (D,D1)-invariant edges. Denote by (i, em) themth entry in the order associated to (i, n). Note that if (i, em) and (i, n) have opposite colorsand the number of points above plus the number of points to the right of (i, em) with thesame color as (i, em) is at most p, then (i, em) is an invariant ≤ p-edge. Similarly, if (i, em)and (i, n) have the same color and the number of points above plus the number of points tothe right of (i, em) with the same color as (i, em) is more than n − 2 − p, then (i, em) is aninvariant ≤ p-edge.

Suppose that the entry (i, n) of M(D) is blue (red).

(i) If (i, e1) is red (blue), then it does not have red entries to its right and it has at mosti − 1 red (blue) entries above. Since i − 1 ≤ p, then (i, e1) is an invariant (≤ i − 1)-edge.Because there are no invariant (≤ i − 2)-edges in row i, it follows that (i, e1) is the uniqueinvariant (i− 1)-edge in row i and thus all i− 1 entries above it are red (blue). Similarly, ifthe (i, e1) is blue (red), then all entries in row i are blue (red) and (i, e1) = (i, i+ 1). Hence(i, e1) has n− i− 1 blue (red) entries to its right and perhaps some other blue (red) entriesabove. Since n− i−1 ≥ n− (p+1)−1 ≥ n−2−p, then (i, e1) is an invariant (≤ i−1)-edge.Because there are no invariant (≤ i − 2)-edges in row i, it follows that (i, e1) is the uniqueinvariant (i− 1)-edge in row i and thus all i− 1 entries above it are red (blue).

For 2 ≤ m ≤ p+ 2− i assume that the entry (i, em′) is an invariant (i− 2 +m′)-edge forevery 1 ≤ m′ ≤ m− 1. Note that i− 1 ≤ i− 2 +m′ ≤ p− 1.

If (i, em) is red (blue), then (i, em′) is red (blue) for every 1 ≤ m′ ≤ m− 1. So (i, em) hasexactly m− 1 red (blue) entries to its right and at most i− 1 red (blue) entries above, thatis, (i, em) is an invariant ≤ (i−2+m)-edge. By hypothesis there is a unique invariant k-edgefor every i − 1 ≤ k ≤ p and among the first m − 1 entries there is exactly one invariantk-edge for each i − 1 ≤ k ≤ i− 2 + (m− 1) = i − 3 +m. So (i, em) is the unique invariant(i − 2 + m)-edge (note that 1 ≤ i − 2 + m ≤ p) and thus all the entries above it are red(blue).

If (i, em) is blue (red), then there are exactly n− i+m blue (red) entries to its right andperhaps some others above it. Since n− i+m ≥ n− i− (p+2− i) = n− p+2, then (i, em)is an invariant ≤ (i− 2 +m)-edge. As before (i, em) must be an invariant (i− 2 +m)-edgeand thus it must have only red (blue) entries above.

We have already determined the unique invariant k-edge for each 1 ≤ k ≤ p. So there areno more invariant ≤ p-edges in row i. For p+ 3− i ≤ m ≤ min{p+ 1, n− i− 1}, we provethat the entry (i, em) has at least p+2−m = min{p+2−m, i−1} red (blue) entries above.

If (i, em) is red (blue), then it has m− 1 red (blue) entries to its right. If (i, em) had lessthan p+ 2−m (note that p+ 2−m ≤ i− 1) red (blue) entries above, then it would be aninvariant ≤ p-edge (because (m− 1) + (p+ 1−m) = p) getting a contradiction.

14

If (i, em) is blue (red), then it has n − i − m blue (red) entries to its right. If (i, em)had less than p + 2 − m red (blue) entries above, then it would have a total of at leastn − i −m + (i − 1)− (p + 1 −m) = n − 2 − p blue (red) entries above or to its right, andthus it would be an invariant ≤ p-edge getting a contradiction.

(ii) The proof is the same as for the case p+ 3− i ≤ m ≤ min{p+ 1, n− i− 1} in (i) aswe only used that the mth entry in that range was not an invariant ≤ p-edge.

Lemma 11. If D is crossing optimal, then for 0 ≤ j ≤ ⌊n/2⌋ − 2 we have

E≤≤k(Dj) = 3(

k+33

)

for all 0 ≤ k ≤ ⌊n/2⌋ − 2− j.

Proof. Since D is crossing optimal, then equality must be achieved in the proof of Theorem7, that is, E≤≤k(D) = 3

(

k+33

)

for all 0 ≤ k ≤ ⌊n/2⌋ − 2. This implies that equality must

be achieved throughout the proof of Theorem 6, in particular, E≤≤k(D1) = 3(

k+23

)

for all

0 ≤ k ≤ ⌊n/2⌋ − 2, which is equivalent to E≤≤k(D1) = 3(

k+33

)

for all 0 ≤ k ≤ ⌊n/2⌋ − 3.

In general, for 0 ≤ j ≤ ⌊n/2⌋ − 2, following the proof of Theorem 6, E≤≤k(Dj) = 3(

k+33

)

for 1 ≤ k ≤ ⌊n/2⌋ − 2 − j implies that E≤≤k−1(Dj+1) = 3(

k+23

)

for 1 ≤ k ≤ ⌊n/2⌋ − 2 − j,

which is equivalent to E≤≤k(Dj+1) = 3(

k+33

)

for 1 ≤ k ≤ ⌊n/2⌋ − 2− j − 1.

Lemma 12. If D is crossing optimal, then in M(D) the mth entry in the order associatedto (i, j) has at least min{j − ⌈n/2⌉ −m, i− 1} entries above with different color than (i, j)for all 1 ≤ m ≤ min{j − ⌊n/2⌋ − 1, j − i− 1}.

Proof. Consider the entry (i, j) of M(D). Because D is crossing optimal, it follows fromLemma 11 that

E≤≤k(Dn−j) = 3(

k+33

)

for all 0 ≤ k ≤ ⌊n/2⌋ − 2− (n− j) = j − 2− ⌈n/2⌉.

Consider row i of Dn−j. (Note that Dn−j has j − 1 rows.) If 1 ≤ i ≤ j − 1− ⌈n/2⌉, thenby Lemma 9 for p = j − 2− ⌈n/2⌉, the 2-page matrix M(Dn−j) satisfies that in row i thereis exactly one (Dn−j, Dn−j+1)-invariant k-edge for each i− 1 ≤ k ≤ j − 2− ⌈n/2⌉ and thereare no (Dn−j, Dn−j+1)-invariant (≤ j − 2− ⌈n/2⌉)-edges. Then by Lemma 10(i) if the entry(i, j) in M(D) (actually in M(Dn−j) but we look at it as a submatrix of M(D)) is blue (red),then the mth entry in the order associated to (i, j) has at least min{j − ⌈n/2⌉ −m, i − 1}red (blue) entries above.

If j − ⌈n/2⌉ ≤ i ≤ j − 1, then by Lemma 9 for p = j − 2 − ⌈n/2⌉, the 2-page matrixM(Dn−j) satisfies that in row i there are no (Dn−j, Dn−j+1)-invariant (≤ j−2−⌈n/2⌉)-edges.Then by Lemma 10(ii) if the entry (i, j) in M(D) is blue (red), then the mth entry in theorder associated to (i, j) has at least j − ⌈n/2⌉ −m = min{j − ⌈n/2⌉ −m, i− 1} red (blue)entries above.

15

Corollary 13. If D is crossing optimal, then for 2 ≤ i ≤ ⌈n/2⌉ and ⌈n/2⌉ + 2 ≤ j ≤ n,each of the first j − ⌈n/2⌉ − 1 entries in the order associated to (i, j) has at least one entryabove with different color than (i, j).

Proof. Let 1 ≤ m ≤ j −⌈n/2⌉− 1. Since ⌊n/2⌋ and i are at most ⌈n/2⌉, then m ≤ min{j −⌊n/2⌋−1, j−i−1}. Alsom ≤ j−⌈n/2⌉−1 and i ≥ 2 imply that max{j−⌈n/2⌉−m, i−1} ≥ 1.Thus by Lemma 12, the mth entry in row i in the order associated to (i, j) has at least oneentry above with different color than (i, j).

Corollary 14. If D is crossing optimal, then for n ≥ 3, 2 ≤ i ≤ ⌊n/2⌋− 1, and ⌈n/2⌉+ i ≤j ≤ n, all entries above the first j − i + 1 − ⌈n/2⌉ entries in the order associated to (i, j)have different color than (i, j).

Proof. Let 1 ≤ m ≤ j − i + 1 − ⌈n/2⌉. Since i ≥ 2 and n ≥ 3, then m ≤ min{j − ⌊n/2⌋ −1, j − i − 1}. Also m ≤ j − i + 1 − ⌈n/2⌉ implies that max{j − ⌈n/2⌉ −m, i − 1} ≥ i − 1.Thus by Lemma 12, the mth entry in row i in the order associated to (i, j) has at least i− 1entries above, (i.e., all entries above it) with different color than (i, j) in M(D).

Lemma 15. Suppose that D is crossing optimal and 0 ≤ k ≤ ⌊n/2⌋−2. Then all ≤ k-edgesof D belong to the union of the first k + 1 rows and the last k + 1 columns of M(D).

Proof. Suppose by contradiction that the entry (i, j) of M(D) represents a k-edge and is notin the first k+1 rows (i ≥ k+2) or in the last k+1 columns (j ≤ n−k−1). SinceD is crossingoptimal, equality must be achieved in Inequality (8) and thus we have that all (D,D1)-invariant ≤ k-edges belong to the first k+1 columns. So (i, j) is not (D,D1)-invariant, thatis, ij is a (k− 1)-edge in D1. Equality in Theorem 6 implies that E≤≤k−1(D1) = 3

(

k+22

)

andas before all (D1, D2)-invariant (≤ k− 1)-edges belong to the last k columns of M(D1), thatis, columns n− k, n− k+ 1, . . . , n− 1 of M(D). So ij is not a (D1, D2)-invariant edge, thatis, ij represents a (k − 2)-edge in D2. In general, assuming that ij is a (k − l)-edge in Dl

and since E≤≤k−l(Dl) = 3(

k−l+32

)

, then all (Dl, Dl+1)-invariant (≤ k − l)-edges belong to thelast k+1− l columns of M(Dl) (i.e., columns n− k, n− k+1, . . . , n− l of M(D)). So (i, j)is not a (Dl, Dl+1)-invariant edge, that is, (i, j) represents a (k − l− 1)-edge in Dl+1. Whenl = k− 1, (i, j) is a 0-edge in M(Dk) that is not in the last column of M(Dk) (column n− kof M(D)). Since there are at least three 0-edges in the first column and row of M(Dk) andi ≥ 2, then E≤≤0(Dk) ≥ 4, but E≤≤0(Dk) must be 3, getting a contradiction.

We extend the standard terminology from the geometrical setting, and call a (⌊n/2⌋−1)-edge a halving edge.

Lemma 16. If D is crossing optimal, then the entries (⌊n/2⌋, ⌈n/2⌉+1), (⌊n/2⌋, ⌊n/2⌋+1),and (⌈n/2⌉, ⌈n/2⌉ + 1) of M(D) are halving edges.

16

Proof. This follows from Lemma 15 as all ≤ (⌊n/2⌋ − 2)-edges of D belong to the union ofthe first ⌊n/2⌋ − 1 rows (top to bottom) and the last ⌊n/2⌋ − 1 columns (left to right) of D.The entries (⌊n/2⌋, ⌈n/2⌉+1), (⌊n/2⌋, ⌊n/2⌋+1), and (⌈n/2⌉, ⌈n/2⌉+1) are not in the first⌊n/2⌋ − 1 rows or in the last ⌊n/2⌋ − 1 columns.

Lemma 16 guarantees that the entry (i, i+ 1) in general, and the entry (i, i+ 2) when nis odd, are halving lines in some drawing equivalent to D. The next result states what thismeans in D. We state it only for 1 ≤ i ≤ ⌊n/2⌋ (but it can be stated for ⌈n/2⌉ ≤ i ≤ n aswell) as it is the only case we explicitly use later in the paper.

Lemma 17. Let 1 ≤ i ≤ ⌊n/2⌋. If D is crossing optimal, then M(D) satisfies that thenumber of blue entries in

{(r, i+ 1) : 1 ≤ r ≤ i− 1} ∪ {(i, c) : i+ 2 ≤ c ≤ i+ ⌈n/2⌉} (9)

∪{(i+ 1, c) : i+ ⌈n/2⌉+ 1 ≤ c ≤ n}

is either ⌊n/2⌋ − 1 or ⌈n/2⌉ − 1. If n is odd, then the number of entries in

{(r, i+ 2) : 1 ≤ r ≤ i− 1} ∪ {(i, c) : i+ 3 ≤ c ≤ i+ ⌈n/2⌉} (10)

∪{(i+ 2, c) : i+ ⌈n/2⌉+ 1 ≤ c ≤ n}

with the same color as the entry (i, i+ 2) is either ⌊n/2⌋ − 1 or ⌊n/2⌋.

Proof. In the strip diagram of D, the entry (i, i + 1) of M(D) corresponds to the en-try (⌊n/2⌋, ⌊n/2⌋ + 1) of M(f i−⌊n/2⌋(D)), see Figure 4 (left). Applying Lemma 16 toM(f i−⌊n/2⌋(D)) and noticing that the entries of M(D) in (9) correspond to the entries aboveplus the entries below the entry (⌊n/2⌋, ⌊n/2⌋ + 1) of M(f i−⌊n/2⌋(D)) gives the result. Theproof of the second part is similar, see Figure 4 (right).

Figure 4: A halving line in a drawing equivalent to D seen in the matrix M(D).

Lemma 18. If D is crossing optimal, then there exists a drawing D′ equivalent to D suchthat in M(D′) the ⌈n/2⌉ entries (1, n), (2, n), . . . , and (⌈n/2⌉, n) are blue and the ⌊n/2⌋ − 1entries (1, ⌈n/2⌉+ 1), (1, ⌈n/2⌉+ 2), . . ., (1, n− 1) are red.

17

Proof. For each integer m, let em be the largest integer such that the last em entries inrow ⌊n/2⌋ of M(fm(D)) have the same color. (These entries are (⌊n/2⌋, n − em + 1), . . .,(⌊n/2⌋, n).) Similarly, let e′m be the largest integer such that the first e′m entries in column⌈n/2⌉+1 ofM(fm(D)) have the same color. (These entries are (1, ⌈n/2⌉+1), . . ., (e′m, ⌈n/2⌉+1).) Let E = max{em, e

′m : m ∈ Z}. We claim that E = ⌈n/2⌉. Indeed, suppose that

E ≤ ⌈n/2⌉ − 1 and without loss of generality assume that E = em0for some integer m0. (If

E = e′m0, start with g(D) instead of D.) Then entry (⌊n/2⌋, n − em0

) has a different colorthan the entries to its right, namely, (⌊n/2⌋, n− em0

+1), . . . , (⌊n/2⌋, n). By Lemma 12 (fori = ⌊n/2⌋ and j = n) the entry (⌊n/2⌋, n−em0

) has at least min{n−⌈n/2⌉−1, ⌊n/2⌋−1} =⌊n/2⌋ − 1 entries above with the same color as (⌊n/2⌋, n − em0

). But this means thate′m0−1+⌊n/2⌋−em0

≥ em0+ 1 = E + 1, a contradiction.

Because E = em0= ⌈n/2⌉, all entries in row ⌊n/2⌋ of M(fm0(D)) are blue. By Lemma

16 all entries in column ⌊n/2⌋+1 of M(fm0(D)) above the entry (⌊n/2⌋, ⌊n/2⌋+1) are red.This implies that D′ = fm0+⌊n/2⌋(D) satisfies the statement.

4.3 The structure of crossing optimal drawings

We are finally ready to investigate the structure of crossing optimal drawings. The nextresult is the workhorse behind Theorems 20 and 23, the main results in this section. Tohelp comprehension, we refer the reader to Figure 5.

Theorem 19. Let n ≥ 6, e = 0 for n even and e = 1 for n odd, and let D be a crossingoptimal 2-page book drawing of Kn. Then there exists a drawing D′ equivalent to D suchthat M(D′) satisfies:

1. for 4+ e ≤ s ≤ ⌊n/2⌋+1 and n+2+ e ≤ s ≤ n+ ⌊n/2⌋+1 the entry (r, s− r) is bluefor all max{1, s− n} ≤ r ≤ (s− 5)/2;

2. for ⌈n/2⌉+ 2 + e ≤ s ≤ n and n+ ⌈n/2⌉+ 2 + e ≤ s ≤ 2n− 2− e the entry (r, s− r)is red for all max{1, s− n} ≤ r ≤ (s− 5)/2 (except for (1, n), which by convention isblue);

3. for n odd, the entries (1, ⌈n/2⌉ + 1) and (⌊n/2⌋, ⌈n/2⌉ + 1) are red, and the entries(2, n) and (⌈n/2⌉, ⌈n/2⌉+ 2) are blue.

Proof. Let

TU(D) = {(r, c) ∈ M(D) : 2 ≤ c ≤ ⌈n/2⌉ , 1 ≤ r ≤ c− 1},

R(D) = {(r, c) ∈ M(D) : ⌈n/2⌉ + 1 ≤ c ≤ n, 1 ≤ r ≤ ⌈n/2⌉}, and

TL(D) = {(r, c) ∈ M(D) : ⌈n/2⌉ + 1 ≤ c ≤ n, ⌈n/2⌉+ 1 ≤ r ≤ c− 1}.

18

Figure 5: The even and odd cases in Theorem 19. The crosses in the odd case representpoints whose color is not fixed.

We shall prove the theorem first for those entries that lie on R(D), then for those that lieon TU (D), and finally for those that lie on TL(D).

The entries in R(D)

We refer the reader to Figure 6. By Lemma 18, we can assume that in M(D)

the entries (1, n), (2, n), . . . , (⌊n/2⌋, n) are blue (11)

(in fact (⌈n/2⌉, n) can also be assumed to be blue but we do not use this fact) and

the entries (1, ⌈n/2⌉ + 1), . . . , (1, n− 1) are red. (12)

Moreover, we can assume that

the entry (2, n− 1) is red. (13)

(If it is blue, then M(h ◦ g(D)) satisfies (11), (12), and (13)).

We now prove that for each r such that 2 ≤ r ≤ ⌊n/2⌋,

the entries (r, ⌈n/2⌉+ 1), (r, ⌈n/2⌉+ 2), . . . , (r, 2⌊n/2⌋ − r + 1) are red (14)

andthe entries (r, 2 ⌈n/2⌉ − r + 2) , (r, 2 ⌈n/2⌉ − r + 3) , . . . , (r, n) are blue. (15)

19

Figure 6: The regions TU(D), R(D), and TL(D).

Observe that if r = 2 and n is even, then (15) only concerns the entry (2, n), whichis blue by (11). (For r = 2 and n odd, (15) is an empty claim.) Thus we only needto take care of the base case r = 2 for (14). Since (by (13)) the entry (2, n − 1) is red,by Corollary 13 the first ⌊n/2⌋ − 2 entries in the order associated to (2, n − 1) have ablue point above. By (12) the only candidates to have blue points above them are the⌈n/2⌉ − 2 entries (2, 3), (2, 4), . . . , (2, ⌈n/2⌉). (Note that the order associated to the entry(i, j) only applies to entries in row j to the left of entry (i, j).) Thus the ⌈n/2⌉ − 2 entries(1, 3), (1, 4), . . . , (1, ⌈n/2⌉) are blue if n is even, and at most one of them, say (1, c1), is redif n is odd. Moreover, by Lemma 8 the entries (2, ⌈n/2⌉ + 1), (2, ⌈n/2⌉ + 2), . . . , (2, n − 2)are red.

For the inductive step, suppose that for some 3 ≤ t ≤ ⌊n/2⌋, each row r with 2 ≤ r ≤t − 1 satisfies the result. We now prove (14) and (15) for r = t. Suppose that the entry(t, 2⌈n/2⌉ − t+ 2) is red. Then by Corollary 13 each of the first ⌈n/2⌉ − t+ 1 entries in theorder associated to (t, 2⌈n/2⌉ − t + 2) has at least one blue entry above. Since the entries(t, ⌈n/2⌉+1), . . . , (t, 2⌊n/2⌋− t+2) have all red above, the only candidates are the ⌈n/2⌉− tentries (t, t+1), (t, t+2), . . . , (t, ⌈n/2⌉) and the entry 2⌊n/2⌋− t+3 = 2⌈n/2⌉− t+1 for oddn. But, by Lemma 8, to be a candidate this last entry should be blue, which is impossiblebecause it would be the first entry in the order associated to (t, 2⌈n/2⌉− t+2) with at mostone blue entry above, contradicting Lemma 12. Since there are not enough candidates, thenthe entry (t, 2⌈n/2⌉ − t + 2) is blue.

Now consider the blue entry (t, n). By Corollary 14 the first ⌊n/2⌋ − t + 1 entries in theorder associated to (t, n) have all entries above them red. The only candidates are (t, c1) if itexists, (t, ⌈n/2⌉+1), . . . , (t, 2⌈n/2⌉−t+1). For n even, there are ⌈n/2⌉−t+1 = ⌊n/2⌋−t+1candidates because (t, c1) does not exists, and thus all of them are red by Lemma 8. For nodd, there are at most 2 more candidates than we need. By Lemma 8 any blue entry (t, c)with c ≥ ⌊n/2⌋+2 is not a candidate. Thus at most two of the last ⌈n/2⌉− t+1 candidates

20

are blue. Suppose that one of the entries (t, ⌈n/2⌉+1), (t, ⌈n/2⌉+2), . . . , (t, 2⌊n/2⌋− t+1)is blue. Then there exists ⌈n/2⌉ + 1 ≤ c ≤ 2⌈n/2⌉ − t such that (t, c) is blue and (t, c + 1)is red. Then (t, c) is the first entry in the order associated to (t, c+ 1) and all entries aboveit are red, contradicting Corollary 13. Thus (14) holds and, by Lemma 8 for (i, j) = (t, n),the rest of (15) holds too.

Note that (14) is vacuous if r = ⌊n/2⌋ and n is odd. On the other hand, we argue that itis possible to assume that

for odd n, the entry (⌊n/2⌋, ⌈n/2⌉+ 1) is red. (16)

Indeed, suppose that it is blue. Then, by Lemma 16, (⌊n/2⌋, ⌈n/2⌉+ 1) is a blue halvingentry with ⌊n/2⌋ − 1 red entries above and thus all ⌊n/2⌋ − 1 entries to its right are blue.Hence, by Lemma 16, (⌊n/2⌋, ⌈n/2⌉) is halving with ⌊n/2⌋ blue entries to its right and thusall ⌊n/2⌋−1 entries above are red. Note that M(f ⌊n/2⌋(D)) satisfies (11), (12), and (13) andits entry (⌊n/2⌋, ⌈n/2⌉+ 1) is red. Then we start with f ⌊n/2⌋(D) instead of D.

We now prove that the version of (15) for r = ⌈n/2⌉ also holds:

the entries (⌈n/2⌉, ⌈n/2⌉+ 2) , (⌈n/2⌉, ⌈n/2⌉+ 3) , . . . , (⌈n/2⌉, n) are blue. (17)

Note that (17) only needs to be proved for odd n, since for even n this is the case r = ⌊n/2⌋in (15). Using (11) and (14) it follows that all the entries above (⌈n/2⌉, ⌈n/2⌉+ 1) are red.By Lemma 16 (⌈n/2⌉, ⌈n/2⌉ + 1) is a halving entry, and so it follows that all the entries toits right are blue. This proves (17).

We now prove that for 2 ≤ r ≤ ⌊n/2⌋ − 1

for odd n, the entry (r, n− r + 1) is red. (18)

Note that (16) is a version of (18) for r = ⌊n/2⌋. Observe that M(f ⌈n/2⌉(D)) satisfies (11)and (12). If (2, n − 1) is red in M(f ⌈n/2⌉(D)), then the diagonal (r, n − r) with 1 ≤ r ≤⌊n/2⌋−1 in M(f ⌈n/2⌉(D)) is red by (14). This corresponds to the diagonal (r, n−r+1) with2 ≤ r ≤ ⌊n/2⌋ in M(D). So now assume that the entry (2, n− 1) is blue in M(f ⌈n/2⌉(D)),which corresponds to (⌊n/2⌋, ⌈n/2⌉ + 2) being blue in M(D). In this case, we can assumethat (1, ⌈n/2⌉) is blue. (Otherwise start with M(h ◦ g ◦ f ⌈n/2⌉(D)) instead of D, whichsatisfies (11), (12), (13), (⌊n/2⌋, ⌈n/2⌉ + 1) is red, and (1, ⌈n/2⌉) is blue.) Now, by Lemma16, (⌊n/2⌋, ⌊n/2⌋ + 1) is a halving entry with ⌊n/2⌋ of the entries in (9) blue, then allothers must be red, i.e., (2, ⌈n/2⌉), (3, ⌈n/2⌉), . . . , (⌊n/2⌋ − 1, ⌈n/2⌉) are red. Assume bycontradiction that (r, n− r + 1) is blue for some 2 ≤ r ≤ ⌊n/2⌋ − 1. Then (r, n− r + 2) isblue, otherwise (r, n− r+1) would be the first entry in the order associated to (r, n− r+2)with no blue entry above, contradicting Corollary 13. But now the red entry (r, ⌈n/2⌉) is the(⌈n/2⌉ − r)th entry in the order associated to the blue entry (r, n) with a blue entry above,contradicting Corollary 14 and proving (18).

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Figure 7: The upper triangle TU(D) for even and odd n in the proof of Theorem 19.

We finally observe that (11), (12), (13), (14), (15), (16), (17), and (18) prove Theorem 19for the entries in R(D).

The entries in TU(D)

We refer the reader to Figure 7. We prove by induction on c that for 1 ≤ c ≤ ⌊12⌈n/2⌉⌋,

the entries (c+ e, ⌈n/2⌉+ 2− c), . . . , (⌊n/2⌋ − c, ⌈n/2⌉+ 2− c) are red, (19)

andthe entries (1, ⌈n/2⌉+ 2− c), . . . , (c− 1− e, ⌈n/2⌉+ 2− c) are blue. (20)

We have proved it for c = 1. Suppose that the result holds for all 1 ≤ c ≤ d − 1 and wenow prove it for c = d. By Lemma 17 for i = ⌈n/2⌉+1− d, and since by (15) the ⌊n/2⌋ − dentries {(i, b) | 2⌈n/2⌉ − i+ 2 ≤ b ≤ i+ ⌈n/2⌉} ∪ {(i+ 1, b) | i+ ⌈n/2⌉ + 1 ≤ b ≤ n} in (9)are blue, then (i, i+ 1) has at most d − 1 + e blue entries above. Suppose by contradictionthat (r, i+ 1) is blue for some d+ e ≤ r ≤ ⌊n/2⌋ − d. Then (r, i+ 1) is the first entry in theorder associated to (r, n− r + 1) and has at most ⌈n/2⌉ − 1 − (⌊n/2⌋ − d)− 1 = d − 2 + eblue entries above. By Lemma 12, (r, i + 1) has at least min{⌊n/2⌋ − r, r − 1} blue entriesabove and thus min{⌊n/2⌋ − r, r− 1} ≤ d− 2 + e. But r− 1 > d− 2 + e because r ≥ d+ e,and ⌊n/2⌋ − r ≥ d > d− 2 + e because r ≤ ⌊n/2⌋ − d. Thus (19) holds for c = d.

Look at (i, i+1) again. The ⌊n/2⌋−1−3e entries {(r, i+1) | d+e ≤ r ≤ i−1−e}∪{(i, b) |i+2+e ≤ b ≤ n−i+1} in (9) are red and thus, by Lemma 17, at most other 4e entries are red.For n even, 4e = 0 and thus (20) holds. For n odd, suppose by contradiction that (d−e, i+1)has a red entry above. We prove that in this case the entries (d− e, i+ 1), (d− e+ 1, i+ 1),and (i− 1, i + 1) are red. Since (d − e, n + 1 − d + e) is red, then by Corollary 14 the first⌊n/2⌋ + 2 − 2d + 2e entries in the order associated to (d − e, n + 1 − d + e) have only blue

22

entries above. If (d− e, i+ 1) were blue, then it would be one of the first two entries in theorder associated to (d− e, n+1− d+ e) with at least one red point above. This means that1 ≥ ⌊n/2⌋+2−2d+2e contradicting that d ≤ ⌊1

2⌈n/2⌉⌋. Thus (d−e, i+1) is red. Similarly,

(d − e + 1, i + 1) cannot be blue as it would be the first entry in the order associated to(d − e + 1, n − d + e), which by Lemma 12 should have at most one red entry above, but(d− e+ 1, i+ 1) has now at least 2 red entries above. Now (i− 1, i+1) is the first entry for(i − 1, n + 2 − i) and, by (19), it has at least ⌈n/2⌉ + 1 − 2d + e red entries above, i.e., atmost d− 2− e blue entries above. But by Lemma 12, the first entry in the order associatedto the red entry (i − 1, n + 2 − i) has at least min{d − 1, i − 2} blue entries above. Thusmin{d−1, i−2} ≤ d−2−e, but d−1 > d−2−e and i−2 > d−2−e because d ≤ ⌊1

2⌈n/2⌉⌋,

getting a contradiction. Hence (i−1, i+1) is red. By Lemma 17 at most ⌊n/2⌋ of the entriesin (10) are red, yet we already have ⌈n/2⌉ red entries (namely, at least the ⌈n/2⌉+1−2d+eabove (i− 1, i+1) mentioned before and the 2d− 2 entries {(i− 1, b) | i+2 ≤ b ≤ n− i+2}to its right), getting a contradiction. Thus (20) holds for c = d.

Now we prove that for 2 ≤ c ≤ ⌈12⌈n/2⌉⌉ + 1,

the entries (1, c), (2, c), . . . , (c− 2− e, c) are blue. (21)

Since (c − 1, c) is one of the ⌊n/2⌋ + 5 − 2c entries in the order associated to the red entry(c − 1, n + 2 − c) (we have shown that the ⌊n/2⌋ − 1 − e entries immediately to the left of(n+ 2− c) are red), then (c− 1, c) has at most one red entry above by Lemma 12. Supposeby contradiction that (r, c) is red for some 1 ≤ r ≤ c− 2 − e. Then (r + 1, c) is blue. Since(r+1, n− r) is red, then by Corollary 14 the first ⌊n/2⌋− 2r entries in the order associatedto (r + 1, n − r) have only blue entries above. But (r + 1, c) is one of the first ⌊n/2⌋ − 2rentries and has the red entry (r, c) above, getting a contradiction.

We finally note that (19), (20), and (21) prove Theorem 19 for the entries in TU(D).

The entries in TL(D)

We refer the reader to Figure 8. Consider f ⌈n/2⌉(D). When n is even, see Figure 8 (left),R(D) and R(f ⌈n/2⌉(D)) are identical and thus our previous arguments show that TU(D)and TU(f

⌈n/2⌉(D)) = TL(D) are identical too, concluding the proof in this case. When n isodd, see Figure 8 (right), R(D) and R(f ⌈n/2⌉(D)) are slightly different: for 2 ≤ r ≤ ⌊n/2⌋the diagonal entries (r, n + 1 − r) are red in R(D) and unfixed in R(f ⌈n/2⌉(D)), and for3 ≤ r ≤ ⌊n/2⌋ the diagonal entries (r, n+2−r) are unfixed in R(D) and blue in R(f ⌈n/2⌉(D)).Also the last row of R(D) is blue and the last row of R(f ⌈n/2⌉(D)) is unfixed. However, thelast column of TU(f

⌈n/2⌉(D)) is red and this is what allows us to mimic the arguments usedfor (19), (20), and (21) to show that TL(D), which corresponds to TU (f

⌈n/2⌉(D)) minus itslast column, satisfies the statement. More precisely, it can be proved by induction on c thatfor 1 ≤ c ≤ ⌊1

2⌈n/2⌉⌋, in M(f ⌈n/2⌉(D))

the entries (c + 1, ⌈n/2⌉+ 1− c), . . . , (⌊n/2⌋ − c− 1, ⌈n/2⌉+ 1− c) are red (22)

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Figure 8: The lower triangle TL(D) versus the upper triangle TU(f⌈n/2⌉(D)) for even and

odd n in the proof of Theorem 19.

andthe entries (1, ⌈n/2⌉+ 1− c), . . . , (c− 2, ⌈n/2⌉+ 1− c) are blue. (23)

We omit the proofs of (22) and (23), as they very closely resemble the proofs of (19) and(20).

Similarly, it can be proved by induction that for 2 ≤ c ≤ ⌈12⌈n/2⌉⌉, in M(f ⌈n/2⌉(D))

the entries (1, c), (2, c), . . . , (c− 3, c) are blue. (24)

The proof of (24) is also omitted, as it very closely resembles the proof of (21).

We finally note that (22), (23), and (24) prove Theorem 19 for the entries in TU(L).

4.4 The number of crossing optimal drawings

Theorem 19 completely determines M(D′) when n is even, which means that in this casethere is essentially only one crossing optimal drawing.

Theorem 20. For n even, up to homeomorphism, there is a unique crossing optimal 2-pagebook drawing of Kn.

24

Proof. The result is easily seen to hold for n = 2 and n = 4. For n ≥ 6 Theorem 19completely determines M(D′). Note that this matrix corresponds to the drawings by Blazekand Koman [5].

In contrast to the even case, for n odd there is an exponential number of non-equivalentcrossing optimal 2-page book drawings of Kn. For any odd integer n ≥ 5, we construct2(n−5)/2 non-equivalent crossing optimal drawings of Kn. In fact, these 2(n−5)/2 drawings arepairwise non-homeomorphic. To prove this, we need the next two results.

Theorem 21. For every n ≥ 13 odd, every crossing optimal 2-page book drawing of Kn hasexactly one Hamiltonian cycle of non-crossed edges, namely the one obtained from the edgeson the spine and the 1n edge.

Proof. Assume n ≥ 13 is odd. To show that 123 . . . n is the only non-crossed Hamiltoniancycle, we show that all other edges are crossed at least once. Assume that D has the formdescribed in Theorem 19. Let (r, c) be an entry of M(D) that does not represent an edgeon the spine or the 1n edge. Let

(r, c)+ =

{

(r + 1, c+ 1) if c < n, or(1, r + 1), if c = n,

and (r, c)− =

{

(r − 1, c− 1) if r > 1, or(c− 1, n), if r = 1.

Note that the edges corresponding to (r, c)+ and (r, c)− cross the edge rc if they have thesame color as (r, c).

First assume that 3 ≤ c − r ≤ n − 3. Suppose that (r, c) is a blue entry specified byTheorem 19. If 5 ≤ r + c ≤ ⌊n/2⌋ − 1 or if n + 3 ≤ r + c ≤ n + ⌊n/2⌋ − 1, then note thatthe entry (r, c)+ is also blue according to Theorem 19, and thus the edges corresponding to(r, c) and (r, c)+ cross each other.

Because n ≥ 13, if ⌊n/2⌋ ≤ r+ c ≤ ⌊n/2⌋+ 1 or n+ ⌊n/2⌋ ≤ r+ c ≤ n+ ⌊n/2⌋+1, then5 ≤ ⌊n/2⌋− 2 ≤ r+ c− 2 ≤ ⌊n/2⌋+1 or n+3 ≤ n+ ⌊n/2⌋− 2 ≤ r+ c− 2 ≤ n+ ⌊n/2⌋+1,respectively. Thus the entry (r, c)− is also blue according to Theorem 19, and thus the edgescorresponding to (r, c) and (r, c)− cross each other.

A similar argument shows that for every red entry (r, c) specified by Theorem 19, either(r, c)+ or (r, c)− is also a red edge.

Second, assume that c− r = n−2, that is (r, c) ∈ {(1, n−1), (2, n)}. If (r, c) = (1, n−1),then (r, c) is red and because (2n− 4) ≥ n+ ⌈n/2⌉+ 2 for n ≥ 13, it follows that rc crossesthe edge corresponding to (n − 3, n), which is red. If (r, c) = (2, n), then (r, c) is blue andbecause ⌊n/2⌋ ≥ 4 for n ≥ 13, it follows that rc crosses the edge corresponding to (1, 4),which is blue.

Suppose now that the color of (r, c) is not determined by Theorem 19. First assume thatr+ c ∈ {⌊n/2⌋+2, ⌈n/2⌉+2, n+ ⌊n/2⌋+2, n+ ⌈n/2⌉+2}. Again, by Theorem 19 note that

25

(r, c)− is blue and (r, c)+ is red. Similarly, if r + c = n + 2, then (r, c)− is red and (r, c)+ isblue. Thus regardless of its color, the edge rc will cross one of the two edges correspondingto these two entries.

Finally, assume c − r = 2. From Theorem 19, the number of red entries of the form(t, r+ 1) or (r+ 1, d), with 1 ≤ t ≤ r and r+ 3 ≤ d ≤ n is at least ⌊n/2⌋ − 5 ≥ 1. A similarstatement holds for the number of blue entries of the same form. Thus there is at least oneblue edge (not on the spine) and at least one red edge incident to r + 1. One of these twoedges will necessarily cross the edge rc regardless of its color.

Note that for n ≤ 11 the above approach cannot guarantee that there are no additionalnon-crossed edges. For example for n = 11 the element (1, 10) cannot be determined.However, these small cases can be handled by exhaustive enumeration, which shows that forcrossing optimal drawings there are no such edges for n = 11 and no alternative Hamiltoniancycles for n = 9. For n = 5, 7 there exist alternative Hamiltonian cycles of non-crossed edges,but they do not lead to additional equivalences between the crossing optimal drawings.

Corollary 22. If D and D′ are crossing optimal 2-page book drawings of Kn, then either Dand D′ are not homeomorphic, or else M(D) and M(D′) are equivalent.

Proof. If n is even the result is trivial by Theorem 20. If n is odd and n ≤ 11, then usingTheorem 19 we exhaustively found all equivalence classes of crossing optimal drawings. Thereare 1, 4, 9, and 25 equivalence classes for n = 5, 7, 9, and 11, respectively. We verified thatall of these equivalence classes were topologically distinct. If n ≥ 13 and D and D′ arecrossing optimal 2-page book drawings, then by the previous theorem both D and D′ haveonly one non-crossed Hamiltonian cycle. Thus if H : D → D′ is a homeomorphism, thenH must send the Hamiltonian cycle 123 . . . n to itself. It follows that H restricted to thiscycle is the composition of a rotation of the cycle with either the identity, or the functionthat reverses the order of the cycle. Moreover, once the edges on the spine are fixed, thedrawing is determined by the colors of the remaining edges. Thus either H is determined byits action on the cycle, or else H switches the blue edges not on the spine with the red edges.In other words, M(D′) = M(H(D)) = M((ha ◦ gb ◦ f i)(D)) for some i ∈ {0, 1, 2, . . . , n− 1}and a, b ∈ {0, 1}. Thus M(D) and M(D′) are equivalent.

Theorem 23. For n odd, there are at least 2(n−5)/2 pairwise non-homeomorphic crossingoptimal 2-page book drawings of Kn.

Proof. As usual let 1, 2, . . . , n be the vertices of Kn. Let rc be an edge of Kn that is noton the Hamiltonian cycle H = 12 . . . n, we color rc red or blue according to the followingrule: if r + c ≡ s (mod n) for some integer 2 ≤ s ≤ (n + 1)/2, then we color rc blue, ifr + c ≡ s (mod n) for some integer (n + 5)/2 ≤ s ≤ n + 1, then we color rc red. Finally, ifr + c ≡ (n+ 3)/2 (mod n), then we color rc either red or blue. See (Figure 9.)

26

Figure 9: The 28 crossing optimal drawings (only 27 non-equivalent) for n = 19 in Theo-rem 23. They are obtained by assigning arbitrary colors to the crosses in this matrix.

We first argue that all of these colorings yield crossing optimal drawings of Kn regardlessof the color of the (n− 3)/2 edges rc for which r + c ≡ (n+ 3)/2 (mod n).

For every 1 ≤ s ≤ n, let Is = {rc edge: rc /∈ H and r + c ≡ s (mod n)}. Note that|Is| = (n − 3)/2 for all s and

⋃ns=1 Is is the complete set of edges not in H . Moreover note

that each Is is a matching of pairwise non-crossing edges.

Let rc be an edge such that r+ c ≡ (n+3)/2 (mod n). Assume without loss of generalitythat r < c. If td is an edge that crosses rc, then t and d are cyclically separated fromr and c; that is, we may assume that r < t < c and d < r or d > c. To facilitate thecase analysis we may assume that the edges that could cross rc are the edges td such thatr < t < c < d < n+ r, with the understanding that d represents the point d−n when d > n.Let C = {td edge: r < t < c < d < n + r} and consider the function T : C → C defined byT (td) = t′d′ where t′ = r + c− t and d′ = r+ c+ n− d. Note that T is well defined becauser < t′ < c < d′ < n+ r and T is one-to-one on C. Moreover, note that

t′ + d′ ≡ r + c+ n + r + c− t− d (mod n)

≡ 2(r + c)− (t + d) (mod n)

≡ (n + 3)− (t+ d) ≡ 3− (t + d) (mod n),

so t+ d ≡ s (mod n) with 2 ≤ s ≤ (n + 1)/2 if and only if t′ + d′ ≡ 3− (t + d) ≡ n+ 3− s(mod n) and (n+ 5)/2 ≤ n+ 3− s ≤ n+ 1. Thus td and T (td) have different colors, whichmeans that C contains as many red edges as blue edges. Hence rc crosses the same numberof edges independently of its color. This shows that all the drawings we have described havethe same number of crossings. Finally, we note that the drawing for which all the arbitraryedges have the same color corresponds to the construction originally found by Blazek andKoman [5] having exactly Z(n) = 1

64(n− 1)2(n− 3)2 crossings. Hence all the other drawings

described are crossing optimal as well.

We now argue that every drawing constructed here is equivalent to exactly one otherdrawing, and thus we have constructed exactly 2(n−5)/2 distinct topological drawings. Let D

27

and D′ be two of the crossing optimal drawings we just constructed and suppose that D andD′ are homeomorphic. By Corollary 22, M(D) and M(D′) are equivalent, thus there existsa transformation F : D → D′ such that F = ha ◦ gb ◦ f i with i ∈ {0, 1, 2, . . . , n − 1} andb, a ∈ {0, 1}. First observe that under f , g, or h, the absolute value difference of the numberof red minus blue edges remains invariant. Thus the drawing D in which all of the edgesin I(n+3)/2 are red can only be homeomorphic to the drawing D′ in which all of those edgesare blue. These two are indeed homeomorphic under the function F = h ◦ g ◦ f (n+1)/2. Nowsuppose that the edges I(n+3)/2 in D and in D′ are not all of the same color. Note that f, g,and h send Im into another Im′ , and if Im is monochromatic (all edges of Im have the samecolor) in D, then Im′ is monochromatic in f(D), g(D), and h(D). Since Im is monochromaticin D if and only if m 6= (n+3)/2, then F must send I(n+3)/2 to itself. If b = 0, rc ∈ I(n+3)/2,and r′c′ is the image of rc under F , then r′ + c′ ≡ r− i+ c− i (mod n). Thus r′ + c′ ≡ r+ c(mod n) if and only if i = 0. Because the edges I1 in D are blue and the edges I1 in h(D)are red, it follows that a = 0 and thus F is the identity. Last, if b = 1, rc ∈ I(n+3)/2, and r′c′

is the image of rc under F , then r′+ c′ ≡ (n+1− (c− i))+(n+1− (r− i)) ≡ 2+2i− (r+ c)(mod n). Thus r′ + c′ ≡ r + c (mod n) if and only if i = (n + 1)/2. Because the edges I1in both D and h(f (n+1)/2(D)) are blue, it follows that a = 1 and thus F = h ◦ g ◦ f (n+1)/2.It can be verified that indeed F (D) is one of the drawings we constructed here, and thusexactly half of the drawings we described are pairwise non-isomorphic.

n drawings n drawings n drawings5 1 17 324 29 389447 4 19 748 31 840649 9 21 1672 33 18028811 25 23 3736 35 38521613 58 25 8208 37 81932815 142 27 17968

Table 1: The number of non-homeomorphic crossing optimal 2-page book drawings of Kn

for odd n, 5 ≤ n ≤ 37.

The above theorem gives a lower bound of 2(n−5)/2 for the number of non-equivalentcrossing optimal drawings. As in the crossing optimal drawings of Theorem 19 there are52(n − 5) entries with non-fixed colors, we get an upper bound of 25(n−5)/2 non-equivalent

crossing optimal drawings. With exhaustive enumeration we have been able to determinethe exact numbers of non-equivalent crossing optimal drawings for n ≤ 37, cf. Table 1. Theobtained results suggest an asymptotic growth of roughly 20.54n, rather close to our lowerbound.

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1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

Figure 10: A 2-page book drawing of K8 with four 0-edges (namely (1, 7), (1, 8), (2, 7),and (2, 8)) and four 1-edges (namely (1, 5) (1, 6), (3, 8), and (4, 8)). This shows that theinequality E≤k(D) ≥ 3

(

k+22

)

, which holds for every geometric drawing D of Kn, does notnecessarily hold if D is a topological drawing.

5 Concluding remarks

It was proved by Abrego and Fernandez-Merchant [1] and by Lovasz et al. [19] that theinequality E≤k (P ) ≥ 3

(

k+22

)

holds (in the geometric setting) for every set P of n points ingeneral position in the plane and for every k such that 0 ≤ k ≤ ⌊n/2⌋ − 2. This inequalityused with the rectilinear version of Theorem 1 gives Z (n) as a lower bound for the rectilinearcrossing number ofKn [1]. In contrast to the rectilinear case, the inequality E≤k (D) ≥ 3

(

k+22

)

does not hold in general for topological drawings D of Kn, not even for general 2-pagedrawings (see Figure 10). This shows the relevance of introducing the parameter E≤≤k (D)(for which Theorem 6 can be established, leading to the 2-page crossing number of Kn).However, the inequality E≤k (D) ≥ 3

(

k+22

)

does hold for crossing optimal 2-page drawingsof Kn. For a proof of this, and other interesting observations on crossing optimal drawingsof Kn, we refer the reader to Section 4 in the proceedings version of this paper [2].

Our approach to determine k-edges in the topological setting is to define the orientationof three vertices by the orientation of the corresponding triangle in a good drawing of thecomplete graph. It is natural to ask whether this defines an abstract order type. To thisend, the setting would have to satisfy the axiomatic system described by Knuth [18]. Butit is easy to construct an example which does not fulfill these axioms, that is, our settingdoes not constitute an abstract order type as described by Knuth [18]. It is an interestingquestion for further research how this new concept compares to the classic order type, bothin terms of theory (realizability, etc.) and applications.

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We believe that the developed techniques of generalized orientation, k-edge for topologicaldrawings, and ≤≤k-edges are of interest in their own. We will investigate their usefulness forrelated problems in future work. For example, they might also play a central role to approachthe crossing number problem for general drawings of complete and complete bipartite graphs.

6 Acknowledgments

O. Aichholzer is partially supported by the ESF EUROCORES programme EuroGIGA,CRP ComPoSe, under grant FWF [Austrian Fonds zur Forderung der WissenschaftlichenForschung] I648-N18. P. Ramos is partially supported by MEC grant MTM2011-22792 andby the ESF EUROCORES programme EuroGIGA, CRP ComPoSe, under grant EUI-EURC-2011-4306. G. Salazar is supported by CONACYT grant 106432. This work was initiatedduring the workshop Crossing Numbers Turn Useful, held at the Banff International ResearchStation (BIRS). The authors thank the BIRS authorities and staff for their support.

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